also help with
Find c>0 such that the area of the region enclosed by the parabolas
y=x^2-c^2 and y=c^2-x^2 is 430.
find the area of the region for
2y=4sqrt(x) , y=3, and 2y+3x=7
3 answers
3 graphs:
y = 2√x
y = 3
y = (-3/2)x + 7
see:
http://www.wolframalpha.com/input/?i=plot++y+%3D+2√x+%2C+y+%3D+3+%2C+y+%3D+%28-3%2F2%29x+%2B+7+%2C+2%3Cx%3C3
intersection of y = (-3/2)x+7 and y = 2√x
is aprr A(2.54, 3.188)
intersection of y = 2√x and y = 3
is B(9/4, 3)
intersection of y = (-3/2)x + 7 and y = 3 is
C(8/3, 3)
So we want the "trianglular" shaped region ABC
From A, draw a perpendicualar to meet y = 3 at P
So APC is a right-angled triangle
area is easy to find
for the other
area = ∫(2√x - 3) dx from x = 9/4 to 2/54
looks routine, messy arithmetic.
y = 2√x
y = 3
y = (-3/2)x + 7
see:
http://www.wolframalpha.com/input/?i=plot++y+%3D+2√x+%2C+y+%3D+3+%2C+y+%3D+%28-3%2F2%29x+%2B+7+%2C+2%3Cx%3C3
intersection of y = (-3/2)x+7 and y = 2√x
is aprr A(2.54, 3.188)
intersection of y = 2√x and y = 3
is B(9/4, 3)
intersection of y = (-3/2)x + 7 and y = 3 is
C(8/3, 3)
So we want the "trianglular" shaped region ABC
From A, draw a perpendicualar to meet y = 3 at P
So APC is a right-angled triangle
area is easy to find
for the other
area = ∫(2√x - 3) dx from x = 9/4 to 2/54
looks routine, messy arithmetic.
nice symmetry, should be easy to see that the x-intercepts are (c,0) and (-c,0) for both
so just double the area from 0 to c
area = 2∫(c^2 - x^2 - (x^2 - c^2)) dx from 0 to c
= 2∫(2c^2 - 2x^2) dx from 0 to c
= 2[2c^2 x - (2/3)x^3] from 0 to c
= 2(c^3 - (2/3)c^3]
= 2(1/3)c^3
= 2/3 c^3
but 2/3 c^3 = 430
c^3 = 645
c = ∛645 = appr 8.64
check my arithmetic
so just double the area from 0 to c
area = 2∫(c^2 - x^2 - (x^2 - c^2)) dx from 0 to c
= 2∫(2c^2 - 2x^2) dx from 0 to c
= 2[2c^2 x - (2/3)x^3] from 0 to c
= 2(c^3 - (2/3)c^3]
= 2(1/3)c^3
= 2/3 c^3
but 2/3 c^3 = 430
c^3 = 645
c = ∛645 = appr 8.64
check my arithmetic